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Waves and Optics Flashcards

Waves, Optics.

1
Q

Define longitudinal wave

A

A wave with a direction of vibration that is parallel to the direction of energy transfer

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2
Q

Define transverse wave

A

A wave with a direction of vibration that is perpendicular to the direction of energy transfer.

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3
Q

Define amplitude of a wave

A

The maximum displacement of the wave from the equilibrium position

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4
Q

Give three examples of waves that are transverse

A
  1. Electromagnetic radiation e.g misrowaves, infrared etc
  2. Surface of water
  3. Rope
  4. s waves
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5
Q

Give three examples of longitudinal waves

A

Sound

Ultrasound

P waves

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6
Q

State the order of the electromagnetic spectrum from lowest frequency/energy to highest frequency/energy

A

Radio

Microwaves

Infrared

Visible

Ultra violet

X - rays

Gamma rays

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7
Q

How do we know waves transfer energy?

A
  1. Electromagnetic waves heats things up when they are absorbed
  2. X-rays and gamma rays knock electrons out of their orbits (ionisation)
  3. Loud sounds cause oscillations of air particles which make things vibrate.
  4. Waves power can be used to generate electricity.
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8
Q

Define wavelength

A

The distance from a point on the wave to the same point on the adjacent wave. It is measured in meters

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9
Q

Define frequency

A

Is the number of waves that pass a point every second. It is measured in hertz (Hz)

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10
Q

Define period

A

It is the time taken for each one whole wave to pass a fixed point. It is measured in seconds (s)

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11
Q

Define phase difference

A

the amount by which one wave lags behind another wave.

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12
Q

What units can phase difference be measured in?

A

Phase and phase difference can be measured in:

  1. degrees,
  2. radians
  3. fractions of a cycle.
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13
Q

Define monochromatic

A

Single wavelength

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14
Q

Define coherent

A

Waves with the same frequency, a constant phase difference and a similar amplitude

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15
Q

Describe an experiment to measure the speed of sound

A
  1. Start with signal generator + 2 microphones in phase.
  2. Move 1 microphone and measure the distance between them.
  3. Measure the time delay using the picoscope
  4. collect a range of distance and time measurements.
  5. Draw a graph of distance against time.
  6. Use the gradient to find the speed of sound.
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16
Q

Define polarisation

A

Restriction of a wave to a single plane

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17
Q

If there are two polarising filters inbetween an observer and light, if one is turned about 360 degrees, what would the observer see?

A

Variation in intensity between max and min Two maxima and two minimum in 360 degree rotation

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18
Q

What waves can be polarised?

A

Transverse

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19
Q

State one application of a polarising filter and a reason for its use

A

Camera/sunglasses - Reduce glare

LCD screens - less power is required

3D glasses - Enhanced viewing experience

Looking at stress/strain/fractures of materials

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20
Q

Light was originally thought to be a longitudinal wave, why do we now know it is transverse?

A

It was discovered light was polsriased by reflection so must be transverse.

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21
Q

Explain why it is important to correctly align the aerial of a TV to receive the strongest signal

A

Transmitted radio waves are often polarised.

Aerial rods must be aligned in the same plane of the wave

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22
Q

Define interference

A

Interference – When two waves arrive at the same point and at the same time, the resultant displacement is given by the algebraic sum of the two individual displacements.

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23
Q

Define constructive interference

A

If two waves meet exactly in phase the amplitudes add up to produce large crests and troughs

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24
Q

Define destructive interference

A

If two waves meet exactly out of phase the amplitudes cancel to produce no crests and troughs

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25
Q

State the difference between a progressive wave and a standing wave.

A

Progressive (travelling wave) – transfer energy from one region to another e.g light and sound

Standing (stationary waves) – Are fixed in space and do not travel. They are caused by the superposition of two identical waves. e.g waves on a guitar string

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26
Q

How are stationary waves formed?

A
  1. Vibrations cause waves travels along the string towards the pulley.
  2. The waves reflect from the pulley and travel back to the vibration generator.
  3. The two waves meet and constructively and destructively interfere.
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27
Q

What are the conditions required for standing waves to be formed?

A

The waves must have the same frequency and amplitude.

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28
Q

Define a node

A

NO DisplacEment – Positive displacement from one wave is cancelled by an equal negative displacement from the other wave. (Destructive interference)

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29
Q

Define an antinode

A

A point of maximum displacement (constructive interference)

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30
Q

For a string that is fixed at both ends what are the rules for nodes and antinodes?

A

There must be a node at each end.

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31
Q

Sketch the first 4 harmonics for a guitar string fixed at both ends.

A

see diagram

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32
Q

How can a guitarist raise the fundamental frequency of vibration in their string?

A
  1. Tighten the string
  2. shorter length of string
  3. Use a string of lower mass per unit length
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33
Q

For a closed pipe what are the rules for nodes and antinodes

A

The closed end of the pipe is always a node, the open end of the pipe is always an antinode

34
Q

Sketch the first 3 harmonics for a closed pipe

A

See diagram - note only odd harmonics are possible

35
Q

For an open pipe what are the rules for nodes and antinodes

A

There must be an antinode at each end

36
Q

Sketch the first 3 harmonics for an open pipe.

A

See diagram

37
Q

Define intensity

A

Energy arriving per second per unit area or

power per unit area

38
Q

Define diffraction

A

The spreading of waves on passing through a gap or near an edge

39
Q

When is diffraction noticeable?

A

When the wavelength is similar in size to the gap/obstacle

40
Q

Describe the intensity pattern for single slit diffraction for red light

A

A bright red central maximum (where the majority of the energy arrives)

Alternating dark and bright fringes either side of the central maximum.

The central maximum is twice the width of the outer bright fringes,

The outer bring fringes have much less intensity than the central maximum.

Peak intensity of each fringe decreases with distance from the centre.

See diagram

41
Q

Describe the diffraction pattern of white light through a single slit?

A
  1. A central bright maximum that is white.
  2. Alternating dark and bright fringes either side of the central maximum.
  3. The outer bright fringes are the spectra of colours from Violet to red. (violet has a shorter wavelength)
  4. This is because white light is a mixture of colours each with a different wavelength.
42
Q

State two ways a diffraction pattern on single slit would change if the slit became narrower

A

Increased diffraction - main peak would be mre spread out - see single slit equation

Lower intensity - less light could get through the smaller slit

43
Q

Using the single slit equation state 2 ways to increase the angle of diffraction

A
  1. Larger wavelength
  2. Smaller gap size
44
Q

For single slit diffraction how does increasing the slit width affect the central maximum?

A
  1. Increasing the slit width will decrease the amount of diffraction so the central maximum is narrower.
  2. As the slit is wider more light can pass through so the intensity of the central maximum will increase.
45
Q

For single slit diffraction how does increasing the wavelength affect the central maximum?

A
  1. Increasing the wavelength increases the amount of diffraction. This means the central maximum is wider and the intensity will be lower.
  2. The outer fringes will also become further apart
46
Q

Explain how Young’s double slit arrangement produces interference fringes

A
  1. Light diffracted at both slits and overlap and interfere
  2. Where path lengths differ by whole number of wavelengths, constructive interference occurs producing a bright fringe
  3. Where path lengths differ by 0.5 x n wavelengths, destructive interference occurs producing a dark fringe
47
Q

Explain how to set up Youngs double slit experiment to ensure the waves passing through both slits are coherent

A

After the light source have a single slit, follwed by the double slits

The first narrow single slit gives wide diffraction to ensure that both second slits are illuminated equally

Path difference to second slits are of constant length giving constant phase difference

48
Q

In Youngs double slit what causes the bright and dark fringes?

A

Bright fringes - waves arriving in phase - constructive interference

Dark fringes - waves arriving out of phase - destructive interference

49
Q

Describe the intesity pattern produced by Young’s double slit experiment.

A

A series of bright and dark fringes.

The bright fringes have equal intensity and are evenly spaced.

See diagram

50
Q

Describe the intensity pattern seen when using white light in Youngs double slit experiment

A

Central fringe would be white

Dark fringes would be either side of white fringe

Side fringes are spectra of colours

51
Q

Using the double slit equation state 3 ways to increase the width between the slits

A
  1. Larger wavelength
  2. Increased distance to screen
  3. Decrease the slit seperation
52
Q

In Youngs double slit experiment what changes would you see if you repalced a blue light source with a red light source.

A

Red has a larger wavelength than blue so:

  1. Central fringe would be wider
  2. Fringer spacing would be larger
53
Q

In Young’s double slit how can the fringe spacing be measured accuratly?

A

Measure across several fringes and divide by the number of fringes.

Measure from the centre of a dark fringe to the centre of anothe rdark fringe. The centre is easier to find on a dark fringe rather than a bright fringe.

54
Q

Use wave theory to explain how the fringe pattern is formed in Youngs double slit experiment

A
  1. Slits act as coherent sources
  2. Waves diffract at slits
  3. Waves superpose
  4. Bright patches: constructive
  5. Dark patches: destructive
55
Q

State and explain the effect of using a laser with a shorter wavelength on the maxima spacing in Young’s double slit

A

Maxima closer together as fringe spacing decreases.

use w=𝜆D/s

56
Q

State two requirements for two light sources to be coherent (2)

A

Same wavelength or frequency.

Constant phase difference

57
Q

What is the role of a diffraction grating?

A

Split light into its seperate wavelengths

58
Q

For a diffraction grating state and explain what happens to angle θ in λ = d sin θ when wavelength decreases?

A

Angle θ gets smaller As path difference gets smaller

59
Q

For a diffraction grating describe the pattern seen with whte light?

A
  1. A white central maximum
  2. Following fringes will be the spectra of colours
  3. From violet to red as violet has a shorter wavelength
60
Q

Describe the pattern seen with red light passing through a diffraction grating

A

Red light is monochromatic so

Red central maximum

Outer orders of diffraction are also red

61
Q

In a diffraction grating how can the maximum number of orders to calculated?

A

Put the angle equal to 90o and then round down to the nearest whole number.

62
Q

Which quantities are changed when monochromatic light passes from air into glass?

A

Speed decreases

wavelength decreases

frequency stays constant

63
Q

What is the name for the part of an optical fibre that is around the core?

A

Cladding

64
Q

What are the two conditions required for TIR?

A

TIR only occurs when ray travels from higher refractive index to lower refractive and as long as the incident angle is greater than the critical angle

65
Q

In fibre optics cables why is absorption a problem?

A

Absorption is where some of the signal’s energy is absorbed by the material of the fibre. This energy loss results in the amplitude of the signal being reduced.

66
Q

In fibre optics cables why is dispersion a problem?

A

Dispersion causes pulse broadening. The received pulse is broader than the initial signal. Broadened pulses can overlap each other leading to information loss.

67
Q

State and explain an advantage of a smaller diameter core

A

Reduce multipath dispersion, which would cause poor resolution

68
Q

Describe the structure of a step-index optical fibre outlining the purpose of the core and cladding

A

Core is transmission medium for electromagnetic waves to progress by total internal reflection.

Cladding has a lower refractive index to allow total internal reflection to occur.

Cladding offers protection from scratching that could lead to light loss

69
Q

State and explain two physical properties of the light produced by a laser which makes it different to that produced by a lamp

A
  1. Monochromatic - Waves of single wavelength
  2. Coherent Waves produced in constant phase
70
Q

What is modal dispersion?

A

Is caused by light rays entering the cable at different angles. This causes them to take different paths and therefore take different amounts of time to travel.

71
Q

How can modal dispersion be reduced?

A

To reduce modal dispersion by using a single mode fibre in which light is only allowed to follow a very narrow path.

72
Q

What is material dispersion

A

Is caused by different amounts of refraction experienced by different wavelengths of light.

This will cause some wavelengths to be slowed down more and take longer to travel down the cable.

73
Q

Discuss two changes to reduce material pulse broadening of white light

A

Monochromatic source so speed is constant

Shorter repeaters so pulse is reformed before significant broadening occurs

74
Q

Blue light has a higher refractive index fibre than red light, explain how this difference results in a change in a pulse of white light by the time it leaves the fibre

A

Blue travels slower than red due to greater refractive index.

Red reaches the end of the fibre before blue leading to material pulse broadening

75
Q

When light goes from a lower to a higher optical density (e.g. air to glass) what is the affect on speed, wavelength, frequency and direction?

A
  1. Speed decreases
  2. Frequency stays the same
  3. Wavelength decreases
  4. It bends towards the normal
76
Q

When light goes from a higher to a lower optical density (e.g. glass to air) what is the affect on speed, wavelength, frequency and direction?

A
  1. Speed increases
  2. Frequency stays the same
  3. Wavelength increases
  4. It bends away from the normal
77
Q

Explain why optical fibres used for communication need to have cladding

A

Keeps signal secure

Keeps most light rays in

78
Q

Why are optical fibre cables made as thin as possible?

A
  1. To reduce multipath/multimode dispersion.
  2. This would cause light of different angles arriving at different times causing pulse broadening.
  3. Reduces data loss Less light is lost so better quality signal
  4. Increased probability of TIR as incident angle less likely to be below the critical angle
79
Q

State some applications of optical fibres

A

Endoscope - improve medical diagnosis/reduce infection

Fibre optics - improve rate of transmission of data

80
Q

Explain how glass cladding around the optical fibre’s core improves the security of data being transmitted through it and give a reason

A
  1. Light doesn’t enter cladding so can’t pass across from one fibre to a neighbouring fibre
  2. Fibres without cladding can allow light to pass between fibres when scratched or linked by moisture
  3. Personal data must be transmitted along fibres where there is no danger of light leakage

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